#include <bits/stdc++.h>
using namespace std;
// O(N^3) DP for JOI Heat (for teaching purposes, NOT for OJ)
const int INF = 1e9;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int L, N;
cin >> L;
vector<int> C(L+2);
for (int i = 1; i <= L; i++) cin >> C[i];
cin >> N;
vector<int> X(N+1);
for (int i = 1; i <= N; i++) cin >> X[i];
// For each road i: store the times of patients
vector<vector<int>> tim(L+2);
for (int t = 1; t <= N; t++) {
int r = X[t];
tim[r].push_back(t);
}
// dp[i][j][t] : after processing road i,
// sent j people to the right on road i,
// and BV i+1 becomes full at time t (t = N+1 => never full)
// dp value = max helicopters
vector<vector<vector<int>>> dp(L+2,
vector<vector<int>>(N+2, vector<int>(N+3, -INF)));
// Base case: before processing any road
dp[0][0][N+1] = 0;
// Process each road i from 1..L-1
for (int i = 1; i <= L-1; i++) {
int Z = tim[i].size(); // number of patients on this road
// Precompute prefix count:
// cnt[k] = number of patients with time <= tim[i][k-1]
vector<int> cnt(Z+1);
for (int k = 1; k <= Z; k++) cnt[k] = k;
// Try all j = number sent right on this road
for (int j = 0; j <= Z; j++) {
// You send j to the right, Z-j to the left
for (int t = 1; t <= N+1; t++) {
int best = dp[i-1][j][t];
if (best < -1e8) continue;
// left hospital = i, right hospital = i+1
// It has capacity C[i] and C[i+1]
// Let’s decide how many left we send to each hospital:
// left_count = Z - j
// right_count = j
// Compute time BV i becomes full:
int left_need = C[i];
int right_need = C[i+1];
int T_left = N+1, T_right = N+1;
if (Z - j >= C[i]) T_left = tim[i][C[i]-1];
if (j >= C[i+1]) T_right = tim[i][j-1];
int full_time = max(T_left, T_right);
// helicopters = number of patients whose time > full_time
int heli = 0;
for (int k = 0; k < Z; k++)
if (tim[i][k] > full_time) heli++;
// Now update dp[i][j][t_new],
// where t_new = T_right (time BV i+1 becomes full)
int t_new = T_right;
dp[i][j][t_new] = max(dp[i][j][t_new], best + heli);
}
}
}
// Final answer = max dp[L-1][0][t]
int ans = 0;
for (int t = 1; t <= N+1; t++)
ans = max(ans, dp[L-1][0][t]);
cout << ans << "\n";
return 0;
}
